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Maps That Take Lines to Circles, in Dimension 4

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Maps That Take Lines to Circles, in Dimension 4 Maps That Take Lines to Circles, in Dimension 4 V. Timorin Abstract. We list all analytic diffeomorphisms between an open subset of the 4-dimen- sional projective space and an open subset of the 4-dimensional sphere that take all line segments to arcs of round circles. These are the following: restrictions of the quaternionic Hopf fibrations and projections from a hyperplane to a sphere from some point. We prove this by finding the exact solutions of the corresponding system of partial differential equations. 1 Introduction Let U be an open subset of the 4-dimensional real projective space RP4 and V an open subset of the 4-dimensional sphere S4. We study diffeomorphisms f : U → V that take all line segments lying in U to arcs of round circles lying in V . For the sake of brevity we will always say in the sequel that f takes all lines to circles. The purpose of this article is to give the complete list of such analytic diffeomorphisms. Remark. Given a diffeomorphism f : U → V that takes lines to circles, we can compose it with a projective transformation in the preimage (which takes lines to lines) and a conformal transformation in the image (which takes circles to circles). The result will be another diffeomorphism taking lines to circles. Example 1. For example, suppose that S4 is embedded in R5 as a Euclidean sphere and take an arbitrary hyperplane and an arbitrary point in R5. Obviously, the pro- jection of the hyperplane to S4 form the given point takes all lines to circles. The restriction of this projection to some open subset of the hyperplane is a diffeomor- phism between this subset and its image. Diffeomorphisms obtained in this way will be called classical projections. Of course, classical projections live in all dimensions, not only 4. arXiv:math/0309053v1 [math.DG] 3 Sep 2003 Example 2. Another example comes from the quaternionic Hopf fibration. Let us recall the definition. Consider the standard projection from the left (resp., right) quaternionic 2-dimensional vector space H2 to the left (resp., right) quaternionic pro- jective line HP1. The latter is identified with S4, a conformal quaternionic coordinate being the ratio of the homogeneous coordinates (in a specified order). Clearly, this projection descends to RP7 — the real projectivization of H2. Thus we obtain a map from RP7 to S4. This is the quaternionic Hopf fibration. It is known (see e.g. [1]) that the quaternionic Hopf fibration takes all lines to circles. Therefore, the same is true for the restriction of it to any 4-dimensional projective subspace of RP7. 1 Our main result is the following Theorem 1 Suppose that an analytic diffeomorphism f between an open set in RP4 and an open set in S4 takes all lines to circles. Then it is either a restriction of a classical projection or a restriction of a (left or right) quaternionic Hopf fibration. Remark. When we say that a map f : U ⊆ RP4 → S4 is a restriction of a classical projection, we mean the following. There is a projective map i from RP4 to R5 defined everywhere on U, a conformal identification j of S4 with a Euclidean sphere in R5 and a central projection π to this Euclidean sphere from some point such that f = j−1 ◦ π ◦ i. Analogously, the statement “f is a restriction of a quaternionic Hopf fibration” has the following meaning. There is a projective map i form RP4 to RP7, a (right or left) quaternionic Hopf fibration π : RP7 → S4 and a conformal identification j of the sphere in the image of f with the sphere in the image of π such that f = j−1 ◦ π ◦ i. Note in particular, that Theorem 1 is insensitive to a projective transformation in the preimage and a conformal transformation in the image. Actually, it is not hard to see that we can always assume j to be fixed from the very beginning. History of the problem. The problem of finding maps that take lines to circles came from nomography (for an introduction to nomography see [2]). G.S. Khovanskii in 1970-s posed the following question: find all diffeomorphisms between open subsets of R2 that take lines to circles, or, in the language of nomography, that transform nomograms with aligned points to circular nomograms. Circular nomograms are more convenient in practice whereas nomograms with aligned points are theoretically easier to deal with. This problem was solved by A.G. Khovanskii [3] who proved that all such diffeomorphisms come from projections of planes in R3 to Euclidean spheres. Actually, his result was stated in a different form: up to a projective transformation in the preimage and a M¨obius transformation in the image there are only 3 such diffeomorphisms, and they correspond to classical geometries. Izadi in [4] extended the results of Khovanskii to the 3-dimensional case. It turned out (see [5]) that Khovanskii’s theorem does not extend to dimension 4. The simplest counterexample is a complex projective transformation which takes lines to circles but does not come from a projection of a hyperplane to a sphere. This is a particular case of example 2 above. In [5] it was proved that all ample enough rectifiable bundles of circles passing through a point in R4 are obtained by means of example 2 (those obtained from example 1 can be obtained from example 2 as well). This may be regarded as a first step in proving Theorem 1. This article completes the proof. Outline of the proof of Theorem 1. Suppose that f : U → V is a diffeomorphism taking all lines to circles. First choose a projective identification of U with a region in R4 and a conformal identification of V with a region in H, the skew-field of quaternions. Then f can be regarded as a quaternion-valued function on U. For any constant vector field α denote by ∂α the Lie derivative along α. Put Aα = ∂αf. The quaternions Aα are best to be understood as components of a quaternion- 2 valued differential 1-form A on U, namely, the differential of f. This form is closed: dA = 0 or, in components, ∂αAβ = ∂βAα for any pair of constant vector fields α and β on U. By Lemma 5.2 from [5], at every point of U there exists a quaternion Bα that depends linearly on α and satisfies one of the equations ∂αAα = BαAα or ∂αAα = AαBα. The products in the right-hand sides are in the sense of quaternions. Assume that the first equation holds everywhere on U (this is not very restrictive: see the end of Section 4). We show in Section 2 that the integrability condition for this equation is 1 1 ∂ B = B B + C(A A + A A + A A ). α β 2 α β 3 α β β α α β where C is a quaternion-valued function. We call it the first integrability condition. Then we use once again that f takes lines to circles, now to deduce that C is real- valued. If C = 0 identically, then the first integrability condition is integrable, and it gives us the maps from example 2. This will be proved in Section 3. Suppose now that C 6= 0. Then we need to write down an integrability condition for the first integrability condition — the second integrability condition. It will be obtained in Section 4. The second integrability condition turns out to be a differential relation between A, B and C only, with no additional parameters. It expresses all first logarithmic derivatives of C in terms of the values of A and B. The fact that C is real imposes a restriction on possible values of A and B. Namely, Bα = p(Aα)+ Aαq where p is a real-valued 1-form and q is a quaternion. Now fix some admissible values of A, B and C at some point of U. We will show in Section 5 that there exists a map from example 1 with the same values of A, B and C at the given point. On the other hand, from the second integrability condition it follows that such analytic map is unique. Thus our map f is from example 1. 2 First integrability condition In this section, we obtain an integrability condition for the equation ∂αAα = BαAα. (1) Since Aα are derivatives, we have ∂αAβ = ∂βAα. From this and from equation (1) it follows immediately that 1 ∂ A = (B A + B A ). (2) α β 2 α β β α An integrability condition for equation (1) is obtained from the equality ∂α∂βAγ = ∂β∂αAγ by expressing derivatives of A with the help of (2). If we introduce a tensor C given in components by 1 C = ∂ B − B B , αβ α β 2 α β 3 then the integrability condition reads as follows: Cαβ Aγ + Cαγ Aβ = CβαAγ + Cβγ Aα. (3) If we apply to equation (3) the transposition of indices β ↔ γ, then we have: ′ Cαβ Aγ + Cαγ Aβ = CγαAβ + CγβAα. (3 ) Take the sum of equations (3) and (3′). It can be written as follows: ′′ (Cβγ + Cγβ)Aα = (2Cαγ − Cγα)Aβ + (2Cαβ − Cβα)Aγ . (3 ) Put β = γ in equation (3). We obtain that (2Cαβ − Cβα)Aβ = CββAα. (4) or, replacing β with γ, ′ (2Cαγ − Cγα)Aγ = CγγAα. (4 ) ′′ Replace 2Cαβ − Cβα and 2Cαγ − Cγα in (3 ) by their expressions obtained from (4) and (4′), respectively. We obtain −1 −1 −1 −1 Cβγ + Cγβ = CββAα(Aβ Aγ )Aα + CγγAα(Aγ Aβ )Aα . (5) Take a point in U.
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